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UNIT 6 : BASIC TRIGONOMETRY WITH TRIANGLES
LESSON
3: ANGLES IN STANDARD POSITION
Definition: An angle is in standard
position if it has its
vertex at the origin and initial arm
along the positive x-axis. The terminal
arm is found by rotating the initial arm about the origin to a terminal
position in one of the 4 quadrants. The
rotation is positive if it is in the counter – clockwise direction and negative
if in the clockwise direction.
Example 1:
Solution:
x = 4 y = 3 r = 5
Example 2: Angles greater than 900
Solution:
x = -5 y = 12 r = 13
Note that sine is positive and cosine and tangent are negative for a second quadrant angle.
Example 3: Complementary angles (angles
which add to 180o).
Using
your calculator, sin 120o =
0.8660254; also sin 60o = 0.8660254
Hence sin 120o
= sin 60o or
sin(120o) =
sin (180o – 120o) = sin 60o
Note that sine is positive in both the first and second quadrants.
Note 60o is the reference(related acute) angle relative to 120o.
Definition:
The reference (related acute) angle is the angle between the terminal arm
and the x-axis.
In the
above example (120o), it is found by subtracting from 180o.
Similarly, cos 150o = -0.8660254; also
cos 30 = 0.8660254
Hence cos 150o
= -cos 300 or
cos(150o) =
-cos (180o – 150o) = -cos 30o
Note that cosine is positive in the first quadrant
and negative in the second quadrant.
Note 30o is the reference
(related acute) angle relative to 150o (180o – 150o
= 30o)
Result: To find the trigonometric ratios of angles between 90o
and 180o, use the following rules:
Examples:
1. Find cos 135o
Solution:
cos 135o = - cos (180o
– 135o)
= - cos 45o
** 45o is reference (related
acute) angle
= - 0.707106781 **
Using your cacluator
2. Find sin 114o.
Solution:
sin 114o = sin(180o
– 114o)
= sin 66o
= 0.913545457
3. Find tan 161o
Solution:
tan 161o = - tan(180o
– 161o)
= - tan 19o
= - 0.344327613
Note that tangent is positive in the
first quadrant and negative in the second
quadrant.
4. Find sin 123o using a calculator
to 5 decimal places.
Solution:
5. Find sec 149o using the
calculator to 5 decimal places.
Solution:
Solution:
Solution:
Since
csc A is positive, < A could be in either the
first or second quadrant. There are two solutions.